JEE PYQ: Electrostatics Question 24
Question 24 - 2020 (05 Sep Shift 1)
A solid sphere of radius $R$ carries a charge $Q + q$ distributed uniformly over its volume. A very small point like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q$. If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then: (assume the remaining portion to be spherical).
(1) $v^2 = y\left[\frac{qQ}{4\pi\varepsilon_0 R^2 ym} + g\right]$
(2) $v^2 = y\left[\frac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]$
(3) $v^2 = 2y\left[\frac{QqR}{4\pi\varepsilon_0(R+y)^3 m} + g\right]$
(4) $v^2 = 2y\left[\frac{qQ}{4\pi\varepsilon_0 R(R+y)m} + g\right]$
Show Answer
Answer: (4)
Solution
Using energy conservation: $\frac{1}{2}mv^2 = mgy + kQq\left(\frac{1}{R} - \frac{1}{R+y}\right) = mgy + \frac{kQq \cdot y}{R(R+y)}$. So $v^2 = 2y\left[g + \frac{qQ}{4\pi\varepsilon_0 R(R+y)m}\right]$.
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